Consider the reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O. In an acid-base tritration, the stoichiometric point happens when 54.32 mL of 0.3561 M NaOH is added to a 67.21 mL portion of H2SO4. What is the [H2SO4]?
This question involves a standard acid-base titration. moles NaOH = c · V = 0.3561 mmol/mL · 54.32 mL = 19.343352 mmol moles H2SO4 = 19.343352 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 9.671676 mmol Hence [H2SO4]= n/V = 9.671676 mmol / 67.21 mL = 0.1439 M The answer to this question is [H2SO4] = 0.1439 M
